A projectile has a range R and time of flight T. If the range is doubled (by increasing the speed of projection, without changing the angle of projection), the time of flight will become

Options

A. `T// sqrt2`
B. `sqrt2T`
C. `T//2`
D. 2T

Explanation

To solve the problem step by step, we will analyze the given information about the projectile motion and derive the required results.

### Step 1: Understand the formulas for range and time of flight
The range \( R \) of a projectile is given by the formula:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
where \( u \) is the initial velocity, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity.

The time of flight \( T \) is given by:
\[ T = \frac{2u \sin(\theta)}{g} \]

### Step 2: Given conditions
We know from the problem that the range is doubled. Therefore, if the original range is \( R \), the new range \( R' \) will be:
\[ R' = 2R \]

### Step 3: Relate the new range to the initial velocity
Since the angle of projection \( \theta \) remains unchanged, we can express the new range in terms of the new initial velocity \( u' \):
\[ R' = \frac{(u')^2 \sin(2\theta)}{g} \]

Setting the new range equal to twice the original range:
\[ \frac{(u')^2 \sin(2\theta)}{g} = 2 \cdot \frac{u^2 \sin(2\theta)}{g} \]

### Step 4: Simplify the equation
Since \( \sin(2\theta) \) and \( g \) are constant and non-zero, we can simplify:
\[ (u')^2 = 2u^2 \]

### Step 5: Solve for the new initial velocity
Taking the square root of both sides gives:
\[ u' = u \sqrt{2} \]

### Step 6: Calculate the new time of flight
Now, we can find the new time of flight \( T' \) using the new initial velocity \( u' \):
\[ T' = \frac{2u' \sin(\theta)}{g} \]

Substituting \( u' = u \sqrt{2} \):
\[ T' = \frac{2(u \sqrt{2}) \sin(\theta)}{g} \]
\[ T' = \sqrt{2} \cdot \frac{2u \sin(\theta)}{g} \]
\[ T' = \sqrt{2} \cdot T \]

### Conclusion
Thus, the new time of flight when the range is doubled is:
\[ T' = \sqrt{2} T \]